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Homework Help: Theory of relativity - proton entering a medium

  1. Mar 2, 2013 #1
    1. The problem statement, all variables and given/known data
    (see attachment)
    Given data:
    1.The speed of light in empty space is ##3 \times 10^8 m/s##.
    2.The mass of the proton is ##1.67 \times 10^{-27} kg## .
    3.The speed of light in water is ##2.3 \times 10^8 m/s##.
    4.When photon emission begins to occur the initial proton, the final proton, and the emitted photon all travel in the same direction.



    2. Relevant equations



    3. The attempt at a solution
    I don't have any knowledge about theory of relativity but I think this question can be done using the info given in the question.

    Initial energy of proton, ##E_i=\sqrt{c^2p^2+m_0^2c^4}##
    Final energy of proton, ##E_f=\sqrt{c^2p_1^2+m_0^2c^4}##
    where ##m_0## is the rest mass of proton, ##p## is the initial momentum of proton and ##p_1## is the final momentum when it enters the water.
    Using conservation of energy,
    [tex]E_i=E_{photon}+E_f[/tex]
    Using conservation of linear momentum,
    [tex]p=p_1+p_2[/tex]
    where ##p_2## is the momentum of emitted photon.
    I can find ##p_2## in terms of ##p## and ##p_1## but I am not sure what I am supposed to do here. I can substitute p_2 in the energy equation but what am I supposed to do with that equation? :confused:

    Any help is appreciated. Thanks!
     

    Attached Files:

    Last edited: Mar 2, 2013
  2. jcsd
  3. Mar 2, 2013 #2
    Anyone?
     
  4. Mar 2, 2013 #3

    haruspex

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    You've not used the equation Eph = cm|p2|.
    Try approximating Ei, Ef on the basis that p << m0c.
     
  5. Mar 2, 2013 #4
    Thanks Haruspex for your reply!
    I used the equation E=cm|p|, took the approximation and substituted them in the energy equation. I ended up with a relation between p and p1.
    [tex]p=2m_o^2c^2c_m-p_1[/tex]
    What am I supposed to do next?
     
  6. Mar 2, 2013 #5

    haruspex

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    That can't be right - it's dimensionally inconsistent. (I get ##p=2m_oc_m-p_1##.)
    How small can p1 be?
     
  7. Mar 2, 2013 #6
    Woops, I forgot to cancel ##m_oc^2## while taking the approximation. :redface:

    smallest value of p1 can be zero?
     
  8. Mar 3, 2013 #7

    haruspex

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    That's what I would assume.
     
  9. Mar 3, 2013 #8
    ok so I plugged in the values. I get ##7.682 \times 10^{-19}## but this is wrong. :-(
     
  10. Mar 3, 2013 #9

    haruspex

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    Having ended up setting p1 to zero, I realise that we didn't need to do the approximation. Try putting p1=0 but not approximating anything. I think you should get 1.86 10-18. Is that better?
     
  11. Mar 3, 2013 #10
    It still says that its incorrect and now I can't try submitting an answer because all my tries are over. :-(
     
  12. Mar 3, 2013 #11

    haruspex

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    And I'm out of ideas - sorry.
     
  13. Mar 3, 2013 #12
    The answers were published today. The answer given is ##3.58 \times 10^{-37}##.
     
  14. Mar 3, 2013 #13

    haruspex

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    Hmm.. that's numerically about the square of what we calculated. Any idea how that result is obtained?
    EDIT: Just ran a check I wish I'd tried before: with my result, the implied initial speed of the proton > c. I guess setting p1=0 is not the way.
     
    Last edited: Mar 3, 2013
  15. Mar 4, 2013 #14
    Then how? :confused:
     
  16. Mar 4, 2013 #15
    My knowledge regarding general relativity is also meagre but the question asks the minimum magnitude of proton's momentum and not the maximum. You should rather choose for maximum value of p1. Its certainly not 0. What is it ?

    Not quite sure of that. You can not be.

    Edit: Aside, I really suspect that approximation will work.

    Ok,ok. A more clear approach will be to eliminate one of the variables(not p) from momentum conservation , and the plug in energy conservation with two unknowns. Just isolate p in terms of p1 or p2 and analyze what value you can set for p1 or p2 for minimum p. What do you get ? Please do not approximate anything!!

    Really, you should get quadratic sort of equation.
     
    Last edited: Mar 4, 2013
  17. Mar 4, 2013 #16

    haruspex

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    p1 is the proton's momentum after the event. The approximation we tried to begin with suggested that the larger that was, the larger the initial momentum needed to be, so it seemed like the smallest p would go with p1=0. Having selected that, it was easy to rerun the calculation without approximating.
    It gives a quartic, which is why I went for approximation. There might be some cancellation (of p-p1), but the algebra gets a bit heavy.
     
  18. Mar 5, 2013 #17
    A quartic ? That will be bit messy. Ok, but you know that approximation was not giving the correct answer, right?

    Can you state that quartic, in terms of p and p1 here? My exams are over on this very day and I am feeling too exhausted to do this. Or rather, we should wait for the OP to post... :rolleyes:
     
  19. Mar 5, 2013 #18

    haruspex

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    It wasn't just wrong, it was out by about 18 orders of magnitude.
    Initial energy of proton, ##E_i=\sqrt{c^2p^2+m_0^2c^4}##
    Final energy of proton, ##E_f=\sqrt{c^2p_1^2+m_0^2c^4}##
    where ##m_0## is the rest mass of proton, ##p## is the initial momentum of proton and ##p_1## is the final momentum when it enters the water.
    Using conservation of energy,
    [tex]E_i=E_{ph} +E_f[/tex]
    Using conservation of linear momentum, with p2 as the momentum of the photon
    [tex]p=p_1+p_2[/tex]
    ##E_{ph} = c_m|p_2|##
    Putting it all together:
    ##\sqrt{c^2p^2+m_0^2c^4} - \sqrt{c^2p_1^2+m_0^2c^4} = c_m(p - p_1)##
    Square both sides, isolate the surd on one side, square again. Actually it's not a quartic, but rather worse. OTOH, there may well be a factor of (p-p1) to take out.
     
  20. Mar 5, 2013 #19
    Yes !! Thanks Haruspex !! :smile:

    You are correct. I solved it further and the factor (p-p1) cancelled out with the calculation. I have eliminated the radicals and have obtained the following equation :

    c4(p+p1)2 + cm4(p-p1)2 - 2c2cm2p2= 2cm2c2p12 + 4cm2mo2c4

    I guess that we will have mess with this equation to find minimum magnitude of p. Since all terms are basically squared, setting p1=0 , in this should do (but it did not ?).

    There must be usage of some mathematical principle to find the magnitude of p1 so as to minimize p.

    No, I think setting p1=0 should do. I get a quadratic equation in p.
    I am sorry. There must be another way.

    Nonetheless, no doubt this is the only way.
     
    Last edited: Mar 5, 2013
  21. Mar 5, 2013 #20
    Setting p1=0 doesn't give the right answer. Oh my, how did you get the patience to do that dirty algebra. :tongue:
     
  22. Mar 5, 2013 #21
    Then rather, open up the equation which I got in my previous post, and write it in the form : ap2+bp+c=0, where c contains all the terms of p1. The the discriminant b2-4ac≥0, should give you some sort of range of values of p1. And that was not very much algebra. Term p-p1 cancelled out as marked by haruspex.
     
  23. Mar 5, 2013 #22
    I tried to differentiate the energy equation (consisting of surds) wrt ##p_1## to find the minimum value of p. I ended up with a quadratic in ##p_1##. Found two values of ##p_1# but none of them gave the right answer when I substituted them in the energy equation.
     
  24. Mar 5, 2013 #23
    Can you show the working? :tongue:

    And please try the discriminant method which I cited in my previous post. I think it will give you the answer. Setting up a maxima or minima is sometimes problematic. Did you analyze d2p/dp12 also ?

    Also if this does not work follow:

    ap2+bp+c=0

    Now use the quadratic formula and isolate p in terms of p1. One root of the two corresponding roots will be minimum value of p. Then from that root you can make p1 self adjust so that p will be surely minimum.

    If all those method also not work (except that you do a careless mistake), then you will have to contact other HHs of PF. This is getting really messy now.
     
    Last edited: Mar 5, 2013
  25. Mar 5, 2013 #24

    TSny

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    I'm not understanding something here. So, please help me out.

    What if you go to the reference frame in which the proton is initially at rest. So, initially, the total energy in this frame is just the rest-mass energy of the proton. How can the proton in this frame emit a photon and conserve energy?

    The only way that I can imagine the proton emitting a photon is if the proton interacts with the background water. But then you would need to include terms for the momentum and energy transferred to the water.
     
  26. Mar 5, 2013 #25
    Not only that, the very concept of "momentum" is controversial for photons moving in a medium. There are two conflicting definitions and over 100 years of debate which one is correct.
     
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